Calculate N: Mass of 1.0mol Blueberries = Number Automobiles

[plstevens]

The mass of an average blueberry is 0.80g and the mass of an automobile is 2200kg.

Find the number of automobiles whose total mass is the same as 1.0mol of blueberries.

N=_______automobiles

 

[pkleinod]

Not enough information. Please supply the molecular formula for a blueberry. (An average blueberry will do.)

 

[NoTime]

Rather unusual application, but I got to like an instructor that puts some thought into their questions.
Try looking up Avogadro's number.

 

[pkleinod]

How is the Avogadro "number" going to help? One mole of blueberries is not
"an Avogadro number of blueberries"! If you have a mixture of elementary
entities, then you have to specify 1)what those entities are and
2) the mole percent of each entity in the mixture. This information is
essential in order to figure out what the mass of a mole of the mixture is.
A blueberry is not an elementary entity, so the mass of the average blueberry is irrelevant.

For example, 1 mole of a mixture containing 78.09 moles per cent of
N₂, 20.95 moles per cent of O₂, 0.93 mole per cent of
Ar and 0.03 mole per cent of CO₂ has a mass of 28.964 grammes.

To calculate an accurate mass of one mole of blueberries, I would
need to know the entities making up the blueberry (e.g. H₂O,
C₆H₁₂O₆, etc.) and the mole per cent of each of these. Assuming a blueberry is mostly water, I would guess 1 mole of the
mixture we call a blueberry would have a mass of about 20 to 30 grammes.
So the number of automobiles would, to a good approximation, be zero.

Potassium cyanide is usually available in the form of pellets having a
mass of about 30 grams. Taking the elementary entity of potassium cyanide
to be KCN, what is the mass of 1 mole of potassium cyanide?
If the pellets were twice as large, would the mass of 1 mole be larger?
The information needed is the molecular weight of KCN, not the mass of
the pellets.

 

[symbolipoint]

This should be no great trouble. A good proper arrangement for the needed values which I give will be easier to see if someone writes it all longhand in normal notation or puts it into a form for TEX or LaTeX.

N=number of cars;

N*(2200 kg/car) = 1 mol bb*(6.022*10^23 bb/mol bb)*(0.80 g/bb)*(1kg/1000 g)

I find no difficulty deriving that arrangement. An Avogadro number of units is 6.022*10^23 units (I might be mildly in error since the accepted value might be 6.03*10^23, or 6.023*10^23).

How is the Avogadro "number" going to help? One mole of blueberries is not
"an Avogadro number of blueberries"! If you have a mixture of elementary
entities, then you have to specify 1)what those entities are and
2) the mole percent of each entity in the mixture. This information is
essential in order to figure out what the mass of a mole of the mixture is.
A blueberry is not an elementary entity, so the mass of the average blueberry is irrelevant.

For example, 1 mole of a mixture containing 78.09 moles per cent of
N₂, 20.95 moles per cent of O₂, 0.93 mole per cent of
Ar and 0.03 mole per cent of CO₂ has a mass of 28.964 grammes.

To calculate an accurate mass of one mole of blueberries, I would
need to know the entities making up the blueberry (e.g. H₂O,
C₆H₁₂O₆, etc.) and the mole per cent of each of these. Assuming a blueberry is mostly water, I would guess 1 mole of the
mixture we call a blueberry would have a mass of about 20 to 30 grammes.
So the number of automobiles would, to a good approximation, be zero.

Potassium cyanide is usually available in the form of pellets having a
mass of about 30 grams. Taking the elementary entity of potassium cyanide
to be KCN, what is the mass of 1 mole of potassium cyanide?
If the pellets were twice as large, would the mass of 1 mole be larger?
The information needed is the molecular weight of KCN, not the mass of
the pellets.

 

[NoTime]

Taking the elementary entity of potassium cyanide
to be KCN ... The information needed is the molecular weight of KCN.

Is the KCN molecule (not atom) an elementary object?
I might note that your question defines the blueberry as a "molecule" with a "molecular" weight 0.80g.

 

[pkleinod]

What I am objecting to is the expression "the mass of 1.0 mole of blueberries",
which, by itself, says nothing as to what this is.
Now I am well aware that many
use the Avogadro constant as a number and think of a mole of objects
as simply that number of objects. (e.g. a mole of base balls, churches, or
average-sized blueberries). This is deplorable because it is not the
way the mole is used in practice in chemistry:

"The mole is the amount of substance of a system which contains as
many elementary entities as there are atoms in 0.012 kilgrammes of
carbon-12. When the mole is used, the elementary entities
must be specified and may be atoms, molecules,
ions, electrons, other particles, or specified groups of such particles."
This is quoted from "Physico-chemical quantitites and units: the grammar
and spelling of physical chemistry (second edition)" by M. L. McGlashan,
(Royal Institute of Chemistry, London, 1971).

The correct usage is illustrated in this book by many examples, a few of which follow:
1 mole of HgCl has a mass equal to 236.04 grammes.
1 mole of Hg₂Cl₂ has a mass equal to 472.08 grammes.
1 mole of Fe_0.91S has a mass equal to 82.88 grammes.

It is worth pointing out that the Avogadro constant is not needed
to compute the mass of a mole of something. To find the mass of a mole of
H₂SO₄ you need only consult a table of atomic
weights.

NoTime asked whether KCN was an elementary object (My post said "entity").

Is the KCN molecule (not atom) an elementary object?
I might note that your question defines the blueberry as a "molecule" with a "molecular" weight 0.80g.

Yes, of course. Note that "elementary" does not mean that the entity is
indivisible; it simply means that the system under consideration can be
regarded as being built up from many instances of this entity (and perhaps
other entities if one is dealing with a mixture). Also, each entity must
be assigned a relative mass (relative to 1/12 the mass of a carbon-12
atom). This relative mass is termed the molecular weight.

So it is not a good idea to define a blueberry as a "molecule" with a
molecular weight of 80 g. First, every individual blueberry would have
to be considered as a different molecule, since they are not instances
of the same elementary entity. And secondly, the molecular weight in this case
would be about 4.8 x 10^23, not 80 g, since molecular weights are
relative masses and hence have no units. If the expression "mole of
blueberries" is to be of any practical use, then it should tell us
the chemical composition of the blueberry (the molecules therein and
their respective mole fractions). One could then compute something
useful such as the number of moles of fructose in one mole of blueberries.

If the original question had been "Estimate the number of automobiles
whose total mass is equal to the mass of 6.02 x 10^23 blueberries", I would have no objections and would agree with the approach taken by Symbolipoint.
But when one says one mole of blueberries, then I, as a chemist, expect
a precise statement as to what that means, and I expect this statement
to conform to the way chemists normally use the mole concept.

 

[NoTime]

"The mole is the amount of substance of a system which contains as
many elementary entities as there are atoms in 0.012 kilgrammes of
carbon-12. When the mole is used, the elementary entities
must be specified and may be atoms, molecules,
ions, electrons, other particles, or specified groups of such particles."
This is quoted from "Physico-chemical quantitites and units: the grammar
and spelling of physical chemistry (second edition)" by M. L. McGlashan,
(Royal Institute of Chemistry, London, 1971).

Well, the blueberry certainly qualifies under --> "specified groups of such particles".
You may not think it very useful, but I would say the point of the exercise is to help people understand that the idea of mole is not constrained to something like a "mole of Au".
Frankly, I think it is one of the better questions I've seen.

 

[pkleinod]

Ah well. We'll just have to agree to disagree!